[LeetCode]Populating Next Right Pointers in Each Node II

Question

Follow up for problem “Populating Next Right Pointers in Each Node”.

What if the given tree could be any binary tree? Would your previous solution still work?

Note:

You may only use constant extra space.
For example,
Given the following binary tree,

         1       /  \      2    3     / \    \    4   5    7

After calling your function, the tree should look like:

         1 -> NULL       /  \      2 -> 3 -> NULL     / \    \    4-> 5 -> 7 -> NULL

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本题难度Hard。

层次递进法

【复杂度】
时间 O(2^h-1) 空间 O(1)

【思路】
本题与[LeetCode]Populating Next Right Pointers in Each Node不一样的地方在于节点可以少儿子或没儿子。还是利用父节点层来链接子节点层，链接子节点时利用cur代表需要链接的节点。在这里需要用一点技巧，就是用fake来处理子节点层的第一个节点。如下图，last指向的是父节点层的节点。

last    1 -> NULL     \ /  \      2 -> 3 -> NULL     父节点层       \    \ fake-> 4 -> 5 -> NULL   子节点层             |            cur

【代码】

public class Solution {    public void connect(TreeLinkNode root) {        //require        TreeLinkNode last=root;        //invariant        while(last!=null){            TreeLinkNode fake=new TreeLinkNode(0);            TreeLinkNode cur=fake;            //遍历父节点层            while(last!=null){                // 尝试链接左节点                if(last.left!=null){                    cur.next=last.left;                    cur=cur.next;                }                // 尝试链接右节点                if(last.right!=null){                    cur.next=last.right;                    cur=cur.next;                }                last=last.next;            }            // 遍历完后，将last移动到子节点层的第一个，准备链接下一层            last=fake.next;        }    }}

参考

[Leetcode] Populating Next Right Pointers in Each Node 二叉树Next指针